Answer:
False.
Explanation:
If you throw a ball into the air, Earth exerts a force on the ball. Also, the ball in the air exerts a force on Earth.
This ultimately implies that, there is a force pair between the earth and the ball.
Answer:
amplitude
Explanation:
When we increase the volume of a sound, the amplitude of the wave is increased. The frequency of a wave is related to its pitch. If the pitch is high, then the frequency of the wave is high. This means that the wave will looked squashed on an oscilloscope trace.
Answer:
8 seconds
Explanation:
From Newton's second law;
Ft = m(v-u)
F = Force applied
t = time taken
v = final velocity
u = initial velocity
20 * t = 32 (9 - 4)
20t = 32 * 5
t = 32 * 5/ 20
t = 8 seconds
Answer:
The correct option is A
Explanation:
From the question we are told that
The mass number is 
Generally the mean radius is mathematically evaluated as

Here
is a constant with a value 
So



To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is




The amplitude of electric field at the receiver is


The amplitude of induced emf by this signal between the ends of the receiving antenna is


Here,
I = Current
= Permeability at free space
c = Light speed
d = Distance
Replacing,


Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V